Problem: Simplify the following expression and state the condition under which the simplification is valid. $r = \dfrac{y^2 - 64}{y + 8}$
Answer: First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = y$ $ b = \sqrt{64} = 8$ So we can rewrite the expression as: $r = \dfrac{({y} + {8})({y} {-8})} {y + 8} $ We can divide the numerator and denominator by $(y + 8)$ on condition that $y \neq -8$ Therefore $r = y - 8; y \neq -8$